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Hydrogen gas, H₂(g), can be produced by reacting methane, CH₄(g), with steam, H₂O(g), at 300 °C in the presence of a suitable catalyst - VCE - SSCE Chemistry - Question 5 - 2023 - Paper 1
Question 5
Hydrogen gas, H₂(g), can be produced by reacting methane, CH₄(g), with steam, H₂O(g), at 300 °C in the presence of a suitable catalyst. The equation for the reaction... show full transcript
Worked Solution & Example Answer:Hydrogen gas, H₂(g), can be produced by reacting methane, CH₄(g), with steam, H₂O(g), at 300 °C in the presence of a suitable catalyst - VCE - SSCE Chemistry - Question 5 - 2023 - Paper 1
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b. ii. Calculate the equilibrium constant, K, for this reaction at 300 °C.
Answer
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Start by defining the initial moles and changes at equilibrium:
- Initial moles:
- CH₄: 25.0 mol
- H₂O: 25.0 mol
- CO₂: 0.00 mol
- H₂: 0.00 mol
- Initial moles:
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Define changes during the reaction:
- Let x be the moles of CH₄ that react. Then,
- At equilibrium:
- CH₄: (25.0 - x)
- H₂O: (25.0 - 2x)
- CO₂: (0.0 + x)
- H₂: (0.0 + 4x)
- At equilibrium:
- Let x be the moles of CH₄ that react. Then,
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Given that at equilibrium the container contains 6.12 moles of H₂: [ 4x = 6.12 \implies x = 1.53 \text{ mol} ]
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Now substitute the value of x into the equilibrium concentrations:
- CH₄: (25.0 - 1.53 = 23.47) mol
- H₂O: (25.0 - 2 \times 1.53 = 21.94) mol
- CO₂: (1.53) mol
- H₂: (6.12) mol
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Calculate the equilibrium concentrations (in Molarity) for a 1000 L container:
- [ [CH_4] = \frac{23.47}{1000} = 0.02347 \text{ M} ]
- [ [H_2O] = \frac{21.94}{1000} = 0.02194 \text{ M} ]
- [ [CO_2] = \frac{1.53}{1000} = 0.00153 \text{ M} ]
- [ [H_2] = \frac{6.12}{1000} = 0.00612 \text{ M} ]
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Finally, plug these values into the equilibrium constant expression: [ K = \frac{(0.00612)^4 (0.00153)}{(0.02347)(0.02194)^2} \approx 1.90 \times 10^{-3} ]